Question

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29.A radial 11-kV distribution feeder connected to the grid system is shown in FIGURE Q4. SUB-Z has undergone protection refurbishment project involving current transformer (CT) size. Overcurrent plug setting is 100% and earthfault plug setting is 10% of the CT ratio for both relays B and C. No grading margin is required between relay A and B. All relays use the Inverse Definite Minimum Time (IDMT) normal inverse curve. Propose the overcurrent and earthfault setting for relays B and C at SUB-Z (up to four decimal places). State all your assumptions. SUB-X GRID Fault level3 phase at SUB-C = 10 kA Fault leveli plase to ground at SUB-C = 5 kA CT = 200/5 A CT = 200/5 SUB-Y B CT = 200/5 OC PS = 100 %: EF PS = 10% OC TMS = 0.4 : EF TMS = 0.4 Curve = IDMT Normal Inverse SUB-Z

Answer 1

Dt of our current relay fauft current Psma Ctratiu X relaycument sulting Relay cunvint Setting 100 % loxin ie SA va se I'CH - 50 Noynal IDMT invece characterist3 20 t - TMS 0.14 PSM PUR-1 -0.44 X OLY = 0.688 sec (50p.B-1 psm=50, t -0.68 Sec SKA Relay ©: - Sarth fauit poctection. fault current pickup for unbalance consent "10) Socy 10 - SOO A We change Pick Up Current 1 SSOA Secondary side custent = 559x2 200 = 13.75A Psm = fault current 4 Pickup current in actual pickup Secondary site = 13 SA 5000 5 50 Relay operating t = tms 0.14 = 1.24see CPSM90? 1 Gempo? Psm=9.09, 6= 1.24gec = 9.09 = 0.48 0.14